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 Wayne June 8, 2005 05:31

Gallalian Invariance

hi, can someone expalin to me what Gallalian Invariance is.

i am a beginer so need a fairly simple answer, as i keep seeing it written when i read about properties of equations.

 Tom June 8, 2005 05:45

Re: Gallalian Invariance

It's nothing more than (a special case of) Newtons first law; i.e. the equivalence of inertial frames. Mathematically this means that the equations of motion are the same in any two (inertial) frames of reference; e.g. make the transformation x -> x + ct, u -> u + c for any constant c then the equations remain unchanged.

 Wayne June 8, 2005 07:30

Re: Gallalian Invariance

sorry for my lack of understanding - but what is an 'inertial' frame?

do u just mean that where ever i look at it from or how ever i am travelling it looks the same???

is stress gallalian invariant? and if so why?

 Tom June 8, 2005 07:44

Re: Gallalian Invariance

An inertial frame is one moving at constant velocity with respect to another frame (the fixed stars in Newtonian mechanics - see any basic school book on physics or applied mathematics).

The stress tensor is invariant under a Gallalean transformation because in a fluid the stresses depend upon the velocity gradient and not the actual velocity (for a Newtonian fluid stress is proportinal to strain).

If the stress tensor was not invariant then the Navier-Stokes equations would no be invariant.

 Jonas Larsson June 8, 2005 08:13

Re: Gallalian Invariance

Check out:

http://en.wikipedia.org/wiki/Galilean_invariance

All laws of classical Newtonian physics should be Galilean invariant. This of course applies to most fluid dynamics laws, including the Navier-Stokes equations. Galelian invariance is often used as a criteria in developing sound physical models for things like turbulence.

For very high speeds, and applications like astrophysics, where clasical mechanics need to be replaced by relativity the laws of physics need no longer be Galilean invariant. Galilean invarance is then replaced by Lorentz invariance... but that is another interesting story. Wikipedia will give you more info if you are interested.

 multi June 8, 2005 09:17

Re: Gallalian Invariance

That is fine Tom I do believe in that.

Say we have a mixture two-phase flow, can we treat the two fluid media in a way to reduce the stress tensor to pressure????

 Tom June 9, 2005 05:44

Re: Gallalian Invariance

I haven't done much work on mixture theory (I'm assuming you are talking about a mixture rather than two fluids separated by a material surface*) - however since in a 1 phase fluid the answer to your question is no in general (the exception being the inviscid limit) then I think it's safe to say the answer is no. In some cases (e.g. inviscid type limit) then it may be possible in special cases to reduce the stress to a (modified) pressure type term.

(*) In this case, in the inviscid limit, the stress is simply the pressure in each phase which you can treat as a single function. this function will be either continuous (no surface tension) or jumps (with surface tension) across the material surface(s).

 multi June 9, 2005 10:44

Re: Gallalian Invariance

This is interesting staff ! Yes about mixture two-phase flow say gas-liquid. So the relative velocity is taken in account i.e. there is an extra equation. In fact I come across thermodynamically compatible systems where the stress tensor reduced to the pressure. During the analysis I find P = P_1 + P_2 using typical EOS for the mixture and for the gas & liquid phases. The interesting side is that the system not biased on the Navier-Stock equations and written in a hyperbolic conservative form. Do believe in this thermodynamically compatible systems……what do you think???

 Tom June 10, 2005 04:51

Re: Gallalian Invariance

If it's not based upon the generalized Navier-Stokes (i.e. the reacting continua equations) then I'm not too sure what you are doing - although I suspect you're using statistical mechanics (which I know very little about). If your approach is a statistical mechanics approach then the ''extra'' stresses are usual hidden leaving only the explicit treatment of the pressure.

In the case of the reacting continua equations you should be able to see which terms in the equations you are approximating and which you are negecting. From what you say about the relative velocities I would assume that you have retained what Truesdell called (if I recall correctly) the "apparrent" stresses. This leaves the two pressure contributions to the stress tensor, which you have, plus the internal/viscous stresses. These internal stresses comprise the normal viscosity of each fluid and the stress of one fluid upon the other. I suspect that your thermodynamically consistent model does not have these terms? (This is probably a reasonable approximation provided you are not interested in boundary effects such as flow separation.

Hope the above ramblings help,

Tom.

P.S. Maybe you should post your question, in more detail, in a new thread - somebody who knows more about what you're doing may reply?

 David Liu June 17, 2005 04:28

Re: Gallalian Invariance

Navier-Stokes equations is of Galilean invariance? The answer is No! You can try that. For example, we assume that one conical pipe flow in the ground accords with NS equation. Then we move this pipe to another inertia frame which moves in a constant speed.

We will found that it will no longer obey NS equation.

 Tom June 17, 2005 06:02

Re: Gallalian Invariance

The Navier-Stokes equations ARE Galilean invariant - it's the boundary conditions that aren't (by changing to a different inertial frame the boundary will appear to move).

 David Liu June 20, 2005 00:58

Re: Gallalian Invariance

For steady flow, NS equation in Vector form:

After changing to another inertia system, V changed, but Grad V, Grad p, Div.T do not change their value in the new system.

So that the above equation is not correct any more. That is, left side is not equal to right side.

Is it right?

Thanks, David

 Tom June 20, 2005 04:23

Re: Gallalian Invariance

No it's incorrect. A steady state solution in one inertial frame will be unsteady in another (the Galilean transform involves time).

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